Download Document

7.6
NON-RIGHT TRIANGLES
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
The Law of Cosines
Law of Cosines: For a triangle with sides a, b, c, and
angle C opposite side c, we have
c2 = a2 + b2 − 2ab cos C
A
c
b
C
B
a
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Proof of the Law of Cosines
Applying the Pythagorean theorem to the right-hand right triangle:
(a − x)2 + h2 = c2 or a2 − 2ax + x2 + h2 = c2.
Applying the Pythagorean theorem to the left-hand triangle, we get
x2 + h2 = b2.
Substituting this result into the previous equation gives
a2 − 2ax + (x2 + h2) = a2 − 2ax + b2 = c2.
But cos C = x/b, so x = b cos C. This gives the Law of Cosines:
a2 + b2 − 2ab cos C = c2.
A
●
h
b
C●
x
c
a− x
● B
a
Triangle used to derive the Law of Cosines
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Application of the Law of Cosines
Example 1
A person leaves her home and walks 5 miles due east and then 3 miles
northeast. How far has she walked? How far away from home is she?
Solution
Destination
She has walked 5 + 3 = 8 miles in total.
One side of the triangle is 5 miles long,
x
while the second side is 3 miles long
3
◦
and forms an angle of 135
5
135° 45°
Home
with the first.
This is because when the person turns northeast, she turns through an angle
of 45◦. Thus, we know two sides of this triangle, 5 and 3, and the angle
between them, which is 135◦. To find her distance from home, we find the
third side x, using the Law of Cosines:

2

  55.213 ,
x  5  3  2  5  3 cos135  34  30 

 2 
2
2
2
x  55.213  7.431 miles
Notice that this is less than 8 miles, the total distance she walked.
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
The Law of Sines
Law of Sines: For a triangle with sides a, b, c opposite
angles A, B, C respectively:
sin A sin B sin C


a
b
c
A
c
b
C
B
a
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Proof of the Law of Sines
We derive the Law of Sines using the same triangle as in the proof of
the Law of Cosines. Since
sin C = h/b and sin B = h/c, we have h = b sin C and h = c sin B.
This means that b sin C = c sin B and
sin B sin C

b
c
sin A sin B
A similar type of argument (Problem 42) shows that
.

a
b
A
b
h
c
C
B
a
Triangle used to derive the Law of Sines
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Application of the Law of Sines
Example 3
An aerial tram starts at a point one half mile from the base of a
mountain whose face has a 60◦ angle of elevation. (See figure.) The
tram ascends at an angle of 20◦.What is the length of the cable from T
to A?
A
c
Solution
T
20◦
a = 0.5 mile
40◦
120◦ 60◦
C
The Law of Cosines does not help us here because we only know the
length of one side of the triangle. We do however know two angles in
this diagram and can determine the third. Thus, we can use the Law
of Sines:
sinA/a = sinC/c or sin 40◦/0.5 = sin 120◦/c
So
c = 0.5 (sin 120◦/sin 40◦) = 0.674.
Functions Modeling Change:
Therefore, the cable from T to A is 0.674 miles.
A Preparation for Calculus,
4th Edition, 2011, Connally
When to Use
the Laws of Cosines and Sines
• When two sides of a triangle and the angle between
them are known the Law of Cosines is useful. It is also
useful if all three sides of a triangle are known.
• The Law of Sines is useful when we know a side and the
angle opposite it and one other angle or one other side.
• The Ambiguous Case: There is a drawback to using the
Law of Sines for finding angles. The problem is that the
Law of Sines does not tell us the angle, but only its sine,
and there are two angles between 0◦ and 180◦ with a
given sine. For example, if the sine of an angle is 1/2, the
angle may be either 30◦ or 150◦.
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally